Question

Predict the sign of \(\Delta S^{\circ}\) and then calculate \(\Delta S^{\circ}\) for each of the following reactions. a. \(2 \mathrm{H}_{2} \mathrm{~S}(g)+\mathrm{SO}_{2}(g) \longrightarrow 3 \mathrm{~S}_{\text {rhombic }}(s)+2 \mathrm{H}_{2} \mathrm{O}(g)\) b. \(2 \mathrm{SO}_{3}(g) \longrightarrow 2 \mathrm{SO}_{2}(g)+\mathrm{O}_{2}(g)\) c. \(\mathrm{Fe}_{2} \mathrm{O}_{3}(s)+3 \mathrm{H}_{2}(g) \longrightarrow 2 \mathrm{Fe}(s)+3 \mathrm{H}_{2} \mathrm{O}(g)\)

Short answer

For the given reactions: a) The sign of ΔS° is negative and the magnitude is 257.8 J/mol K. b) The sign of ΔS° is positive and the magnitude is 187.0 J/mol K. c) The sign of ΔS° is positive and the magnitude is 36.2 J/mol K.

Step-by-step solution

Observing reaction components' phases#for a Look at the given reaction: \[2H_2S(g) + SO_2(g) \longrightarrow 3S_{\text{rhombic}}(s) + 2H_2O(g)\] Observe the phases of the reactants and the products. We can see that two gaseous substances are reactants, and a solid and a gaseous substance are the products.

Identifying changes in phases#for a There's a decrease in the number of gaseous molecules on the right-hand side of the equation, although the overall number of gaseous species remains the same.

Predicting ΔS° sign#for a Since the gaseous phase exhibits more disorder than the solid phase, the entropy change for this reaction will be negative because the products have more order than the reactants. \[\Delta S^{\circ} < 0\] #a. Calculating ΔS°#

Recall the formula#for a To calculate ΔS°, we need the standard entropies of all substances involved in the reaction. We will use the formula: \[\Delta S^{\circ} = \sum S^{\circ}_{products} - \sum S^{\circ}_{reactants}\]

Using standard entropies#for a From the standard entropy tables, we can find the following values: \(S^{\circ}(H_2S) = 206.1 \frac{J}{mol \cdot K}, S^{\circ}(SO_2) = 248.3 \frac{J}{mol \cdot K}, S^{\circ}(S_{\text{rhombic}}) = 31.8 \frac{J}{mol \cdot K}, S^{\circ}(H_2O) = 188.8 \frac{J}{mol \cdot K}\). Plug these values into the formula: \[\Delta S^{\circ} = [3(31.8) + 2(188.8)] - [2(206.1) + 1(248.3)]\]

Calculating ΔS° value#for a Calculate the entropy change: \[\Delta S^{\circ} = (-257.8) \frac{J}{mol \cdot K}\] For reaction a, the ΔS° is negative, as predicted, and the magnitude is 257.8 J/mol K. #b. Predicting sign of ΔS°#

Observing reaction components' phases#for b Look at the given reaction: \[2SO_3(g) \longrightarrow 2SO_2(g) + O_2(g)\] Observe the phases of the reactants and the products. We can see that all substances involved are in the gaseous phase.

Identifying changes in phases#for b There's an increase in the number of gaseous molecules, the gaseous species number changes from 2 to 3, on the right-hand side of the equation.

Predicting ΔS° sign#for b Since there are more gaseous molecules when the reaction proceeds, the entropy change for this reaction will be positive because the products have more disorder than the reactants. \[\Delta S^{\circ} > 0\] #b. Calculating ΔS°#

Using standard entropies for b From the standard entropy tables, we can find the following values: \(S^{\circ}(SO_3) = 256.9 \frac{J}{mol \cdot K}, S^{\circ}(SO_2) = 248.3 \frac{J}{mol \cdot K}, S^{\circ}(O_2) = 205.2 \frac{J}{mol \cdot K}\). Plug these values into the formula: \[\Delta S^{\circ} = [2(248.3) + 1(205.2)] - [2(256.9)]\]

Calculating ΔS° value#for b Calculate the entropy change: \[\Delta S^{\circ} = (187.0) \frac{J}{mol \cdot K}\] For reaction b, the ΔS° is positive, as predicted, and the magnitude is 187.0 J/mol K. #c. Predicting sign of ΔS°#

Observing reaction components' phases#for c Look at the given reaction: \[Fe_2O_3(s) + 3H_2(g) \longrightarrow 2Fe(s) + 3H_2O(g)\] Observe the phases of the reactants and the products. We can see that solid and gaseous substances are involved in both the reactants and the products.

Identifying changes in phases#for c There's an increase in the number of gaseous molecules on the right-hand side of the equation, although they have the same number of gaseous species.

Predicting ΔS° sign#for c Since the gaseous phase exhibits more disorder than the solid phase, the entropy change for this reaction will be positive because the products have more disorder than the reactants. \[\Delta S^{\circ} > 0\] #c. Calculating ΔS°#

Using standard entropies#for c From the standard entropy tables, we can find the following values: \(S^{\circ}(Fe_2O_3) = 87.4 \frac{J}{mol \cdot K}, S^{\circ}(H_2) = 130.7 \frac{J}{mol \cdot K}, S^{\circ}(Fe) = 27.3 \frac{J}{mol \cdot K}, S^{\circ}(H_2O) = 188.8 \frac{J}{mol \cdot K}\). Plug these values into the formula: \[\Delta S^{\circ} = [2(27.3) + 3(188.8)] - [1(87.4) + 3(130.7)]\]

Calculating ΔS° value#for c Calculate the entropy change: \[\Delta S^{\circ} = (36.2) \frac{J}{mol \cdot K}\] For reaction c, the ΔS° is positive, as predicted, and the magnitude is 36.2 J/mol K.

Key concepts

Thermodynamics

Thermodynamics is the branch of physics that deals with the relationships between heat and other forms of energy. In the context of chemical reactions, it helps us understand how energy is exchanged and how systems move towards equilibrium. One of the key concepts here is entropy, which is a measure of disorder or randomness in a system.

• Entropy ( \(S\) ) tends to increase in natural processes, meaning systems move towards more disorder.
• The Second Law of Thermodynamics states that the total entropy of an isolated system will always increase over time.

For chemical reactions, we are interested in the change in entropy ( \(\Delta S\) ) which indicates how the disorder changes from reactants to products.

• If \(\Delta S^{\circ} > 0\) , the products are more disordered, and the reaction might be driven by entropy to proceed spontaneously.
• If \(\Delta S^{\circ} < 0\) , the products are more ordered.

Understanding \(\Delta S^{\circ} \) helps predict whether a reaction is spontaneous under standard conditions, considering it along with other thermodynamic factors, such as enthalpy.

Phase Transitions

Phase transitions involve a change in the state of matter of a substance, such as from solid to liquid. These transitions play a significant role in thermodynamics and entropy changes. During a phase transition, the energy and entropy can change significantly.

• Each state of matter (solid, liquid, gas) has a different entropy level, with gases generally having the highest entropy due to increased particle movement and disorder.
• Transitions such as melting, vaporization, and sublimation generally result in an increase in entropy \((\Delta S^{\circ} > 0)\) because the systems move towards greater disorder.
• Reactions involving gas to liquid or liquid to solid transitions generally show a decrease in entropy \((\Delta S^{\circ} < 0)\) as they move towards a more ordered state.

In the context of the problems given, understanding phase transitions helps predict the sign of the entropy change based on the direction of the phase change.

Standard Entropy Values

Standard entropy values are important for calculating \(\Delta S^{\circ} \) in chemical reactions under standard conditions (1 bar of pressure and 298 K temperature). These values represent the absolute entropy of a substance in the standard state and are often tabulated for convenience.

• Each element or compound has a unique standard entropy value measured in \(\text{J/mol·K}\) .
• These values depend on the molecular structure, physical state, and temperature of the substance.

Calculating the entropy change of a reaction involves summing the standard entropies of the products and subtracting the sum of the standard entropies of the reactants:
\[ \Delta S^{\circ} = \sum S^{\circ}_{\text{products}} - \sum S^{\circ}_{\text{reactants}} \] Using standard entropy values allows us to predict whether a reaction might proceed spontaneously by examining the disorder associated with the transformation from reactants to products.